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2 years ago

Does a ball ever bounce back to it's drop height?

Physics

2 answers:

Valentin [98]2 years ago

5 0

Answer: No

Explanation: Once you bounce the ball it will go higher and then it will slowly come back down

ELEN [110]2 years ago

3 0

Answer:

depends the material

Explanation:

if its rubber yes for sure

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The greater the mass of an object, the greater its force due to gravity

Serga [27]

What is the question?

6 0

3 years ago

At resonance, what is impedance of a series RLC circuit? less than R It depends on many other considerations, such as the values

denis23 [38]

Answer:

at resonance impedence is equal to resistance and quality factor is dependent on R L AND C all

Explanation:

we know that for series RLC circuit impedance is given by

$Z=\sqrt{R^2+\left ( X_L-X_C \}right )^2$

but we know that at resonance $X_L=X_C$

putting $X_L=X_C$ in impedance formula , impedance will become

Z=R so at resonance impedance of series RLC is equal to resistance only

now quality factor of series resonance is given by

$Q=\frac{\omega L}{R}=\frac{1}{\omega CR}=\frac{1}{R}\sqrt{\frac{L}{C}}$ so from given expression it is clear that quality factor depends on R L and C

3 0

3 years ago

When an object is in circular motion it is constantly changing its velocity.

EleoNora [17]

Its tangential speed is constant although its velocity is changing. As the object changes direction, it results in a changing of positive and negative signs of the velocity. Although, the magnitude of the velocity (speed) is not changing.

8 0

3 years ago

An astronaut has a mass of 75 kg and is floating in space 500 m from his 125,000 kg spacecraft. What will be the force of gravit

nikdorinn [45]

Answer:

1. 2.5×10¯⁹ N

2. 3.33×10¯¹¹ m/s²

Explanation:

1. Determination of the force of attraction.

Mass of astronaut (M₁) = 75 Kg

Mass of spacecraft (M₂) = 125000 Kg

Distance apart (r) = 500 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force of attraction (F) =?

The force of attraction between the astronaut and his spacecraft can be obtained as follow:

F = GM₁M₂ /r²

F = 6.67×10¯¹¹ × 75 × 125000 / 500²

F = 2.5×10¯⁹ N

Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N

2. Determination of the acceleration of the astronaut.

Mass of astronaut (m) = 75 Kg

Force (F) = 2.5×10¯⁹ N

Acceleration (a) of astronaut =?

The acceleration of the astronaut can be obtained as follow:

F = ma

2.5×10¯⁹ N = 75 × a

Divide both side by 75

a = 2.5×10¯⁹ / 75

a = 3.33×10¯¹¹ m/s²

Thus, the acceleration the astronaut is 3.33×10¯¹¹ m/s²

5 0

2 years ago

Point P and point charge Q are separated by a distance R. The electric field at point P has magnitude E. How could the magnitude

Angelina_Jolie [31]

<span>We know , E = kQ/r^2 where q = charge and r is separation between point and point charge. Now, At P, E= kQ/r^2 Since, Q can't be changed, we can do that by varying r 2E = 2kq/r^2 2E = kq/ (r/ sqrt2)^2 Hence, if we bring Q closer such that distance between P and Q becomes r/ sqrt 2, E will get doubled.</span>

5 0

3 years ago